# Find the Missing Number in a sorted array

• Difficulty Level : Easy
• Last Updated : 06 May, 2022

Given a list of n-1 integers and these integers are in the range of 1 to n. There are no duplicates in list. One of the integers is missing in the list. Write an efficient code to find the missing integer.

Examples:

Input : arr[] = [1, 2, 3, 4, 6, 7, 8]
Output : 5

Input : arr[] = [1, 2, 3, 4, 5, 6, 8, 9]
Output : 7

Naive approach: One Simple solution is to apply methods discussed for finding the missing element in an unsorted array. Time complexity of this solution is O(n).

Efficient approach: It is based on the divide and conquer algorithm that we have seen in binary search, the concept behind this solution is that the elements appearing before the missing element will have ar[i] – i = 1 and those appearing after the missing element will have ar[i] – i = 2.

Below is the implementation of the above approach:

## C++

 `// A binary search based program to find the``// only missing number in a sorted array of``// distinct elements within limited range.``#include ``using` `namespace` `std;` `int` `search(``int` `ar[], ``int` `size)``{``    ``// Extreme cases``    ``if` `(ar != 1)``        ``return` `1;``    ``if` `(ar[size - 1] != (size + 1))``        ``return` `size + 1;` `    ``int` `a = 0, b = size - 1;``    ``int` `mid;``    ``while` `((b - a) > 1) {``        ``mid = (a + b) / 2;``        ``if` `((ar[a] - a) != (ar[mid] - mid))``            ``b = mid;``        ``else` `if` `((ar[b] - b) != (ar[mid] - mid))``            ``a = mid;``    ``}``    ``return` `(ar[a] + 1);``}` `int` `main()``{``    ``int` `ar[] = { 1, 2, 3, 4, 5, 6, 8 };``    ``int` `size = ``sizeof``(ar) / ``sizeof``(ar);``    ``cout << ``"Missing number:"` `<< search(ar, size);``}` `// This code is contributed by Pushpesh Raj`

## Java

 `// A binary search based program``// to find the only missing number``// in a sorted array of distinct``// elements within limited range.``import` `java.io.*;` `class` `GFG {``    ``static` `int` `search(``int` `ar[], ``int` `size)``    ``{``        ``// Extreme cases``        ``if` `(ar[``0``] != ``1``)``            ``return` `1``;``        ``if` `(ar[size - ``1``] != (size + ``1``))``            ``return` `size + ``1``;` `        ``int` `a = ``0``, b = size - ``1``;``        ``int` `mid = ``0``;``        ``while` `((b - a) > ``1``) {``            ``mid = (a + b) / ``2``;``            ``if` `((ar[a] - a) != (ar[mid] - mid))``                ``b = mid;``            ``else` `if` `((ar[b] - b) != (ar[mid] - mid))``                ``a = mid;``        ``}``        ``return` `(ar[a] + ``1``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `ar[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``8` `};``        ``int` `size = ar.length;``        ``System.out.println(``"Missing number: "``                           ``+ search(ar, size));``    ``}``}` `// This code is contributed``// by inder_verma.`

## Python3

 `# A binary search based program to find``# the only missing number in a sorted``# in a sorted array of distinct elements``# within limited range`  `def` `search(ar, size):``   ``# Extreme cases``    ``if``(ar[``0``] !``=` `1``):``        ``return` `1``    ``if``(ar[size``-``1``] !``=` `(size``+``1``)):``        ``return` `size``+``1` `    ``a ``=` `0``    ``b ``=` `size ``-` `1``    ``mid ``=` `0``    ``while` `b > a ``+` `1``:``        ``mid ``=` `(a ``+` `b) ``/``/` `2``        ``if` `(ar[a] ``-` `a) !``=` `(ar[mid] ``-` `mid):``            ``b ``=` `mid``        ``elif` `(ar[b] ``-` `b) !``=` `(ar[mid] ``-` `mid):``            ``a ``=` `mid``    ``return` `ar[a] ``+` `1`  `# Driver Code``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``8``]``n ``=` `len``(a)` `print``(``"Missing number:"``, search(a, n))` `# This code is contributed``# by Mohit Kumar`

## C#

 `// A binary search based program``// to find the only missing number``// in a sorted array of distinct``// elements within limited range.``using` `System;` `class` `GFG {``    ``static` `int` `search(``int``[] ar, ``int` `size)``    ``{``        ``// Extreme cases``        ``if` `(ar != 1)``            ``return` `1;``        ``if` `(ar[size - 1] != (size + 1))``            ``return` `size + 1;` `        ``int` `a = 0, b = size - 1;``        ``int` `mid = 0;``        ``while` `((b - a) > 1) {``            ``mid = (a + b) / 2;``            ``if` `((ar[a] - a) != (ar[mid] - mid))``                ``b = mid;``            ``else` `if` `((ar[b] - b) != (ar[mid] - mid))``                ``a = mid;``        ``}``        ``return` `(ar[a] + 1);``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main(String[] args)``    ``{``        ``int``[] ar = { 1, 2, 3, 4, 5, 6, 8 };``        ``int` `size = ar.Length;``        ``Console.WriteLine(``"Missing number: "``                          ``+ search(ar, size));``    ``}``}` `// This code is contributed``// by Arnab Kundu`

## PHP

 ` 1)``    ``{``        ``\$mid` `= (int)((``\$a` `+ ``\$b``) / 2);``        ``if` `((``\$ar``[``\$a``] - ``\$a``) != (``\$ar``[``\$mid``] -``                                   ``\$mid``))``            ``\$b` `= ``\$mid``;``        ``else` `if` `((``\$ar``[``\$b``] - ``\$b``) != (``\$ar``[``\$mid``] -``                                        ``\$mid``))``            ``\$a` `= ``\$mid``;``    ``}``    ``return` `(``\$ar``[``\$a``] + 1);``}` `// Driver Code``\$ar` `= ``array``(1, 2, 3, 4, 5, 6, 8 );``\$size` `= sizeof(``\$ar``);``echo` `"Missing number: "``,``     ``search(``\$ar``, ``\$size``);` `// This code is contributed by ajit.``?>`

## Javascript

 ``

Output

`Missing number:7`

Time Complexity: O(log(N)), where N is the length of given array
Auxiliary Space: O(1)

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